Question

Initially, 1.00 mol of an ideal monatomic gas has 75.0 J of thermal energy. Part A If this energy is increased by 22.0 J , what is the change in entropy?

Answer 1

Answer:

Δ S = 2.51 × 10²³ J/K

Explanation:

change of entropy formula

$\Delta S = \dfrac{3}{2}Nln\dfrac{E_x}{E_n}$

N = 6.02 × 10²³ atoms

Initial thermal energy = Eₙ = 75 J

Final thermal energy,

Eₓ = 75 J  + 22 J

Eₓ = 97 J

$\Delta S = \dfrac{3}{2}Nln\dfrac{E_x}{E_n}$

$\Delta S = \dfrac{3}{2}\times 6.02\times 10^{23} ln\dfrac{97}{75}$

Δ S = 2.51 × 10²³ J/K